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irakobra [83]
3 years ago
11

Can someone plz help? Geometry ‼️

Mathematics
2 answers:
Lilit [14]3 years ago
5 0

Answer:

16

Step-by-step explanation:

Mars2501 [29]3 years ago
4 0
The answer to your question is 16
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7+2x/3 =5 what’s the answer
Jobisdone [24]
<h3>Answer:</h3>

4

<h3>Step-by-step explanation:</h3>

\sf  \frac{7 + 2x}{3}  = 5

\sf  = 7 + 2x = (5 \times 3)

\sf  = 7 + 2x = 15

\sf  = 2x = 15 - 7

\sf  = 2x = 8

\sf  =  \:  \:  \: x = 8 \div 2 \\  \sf x = 4

3 0
2 years ago
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Fred went to his local zoo that featured 6 whale exhibits. If the zoo features 15 exhibits in total, then what percent of the zo
lorasvet [3.4K]

Answer:40%

Step-by-step explanation:

Given

The zoo featured 6 whales exhibits

There are 15 exhibits in total

Percentage of Zoo's exhibits featuring whales

\Rightarrow \dfrac{6}{15}\times 100=40\%

7 0
3 years ago
Which are true about Figure 1 and Figure 2? Select all that apply.
iogann1982 [59]

Answer:

The figures have the same shape, Corresponding angles have the same measure, Corresponding sides have the same ratio, and The figures are similar.

3 0
3 years ago
Write the fact family for 5,8,and 40
Juli2301 [7.4K]
A fact family is basically just different equations with the same numbers getting the same/different numbers as answers you might say... 
this will make more sense: 
5x8=40
8x5=40
40/8=5
40/5=8
4 0
3 years ago
Find the probability of the following events , when a dice is thrown once:
Fudgin [204]

Answer:

Step-by-step explanation:

s a die is rolled once, therefore there are six possible outcomes, i.e., 1,2,3,4,5,6.

(a) Let A be an event ''getting a prime number''.

Favourable cases for a prime number are 2,3,5,

i.e., n(A)=3

Hence P(A)=n(A)n(S)=36=12

(b) Let A be an event ''getting a number between 3 and 6''.

Favourable cases for events A are 4 or 5.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(c) Let A be an event ''a number greater than 4''.

Favourable cases of events A are 5, 6.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(d) Let A be the event of getting a number at most 4.

∴ A={1,2,3} ⇒ n(A)=4,n(S)=6

∴ Required probability =n(A)n(S)=42=23

(e) Let A be the event of getting a factor of 6.

∴ A={1,2,36} ⇒ n(A)=4,n(A)=6

∴ Required probability =46=23

(ii) Since, a pair of dice is thrown once, so there are 36 possible outcomes. i.e.,

(a) Let A be an event ''a total 6''. Favourable cases for a total of 6 are (2,4), (4,2), (3,3), (5,1), (1,5).

i.e., n(A)=5

Hence P(A)=n(A)n(S)=536

(b) Let A be an event ''a total of 10n. Favourable cases for total of 10 are (6,4), (4,6), (5,5).

i.e., n(A)=5

P(A)=n(A)n(S)=336=112

(c) Let A be an event ''the same number of the both the dice''. Favourable cases for same number on both dice are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

i.e., n(A)=6

P(A)=n(A)n(S)=636=16

(d) Let A be an event ''of getting a total of 9''. Favourable cases for a total of 9 are (3,6), (6,3), (4,5), (5,4).

i.e., n(A)=4

P(A)=n(A)n(S)=436=19

(iii) We have, n(S) = 36

(a) Let A be an event ''a sum less than 7'' i.e., 2,3,4,5,6.

Favourable cases for a sum less than 7 ar

7 0
3 years ago
Read 2 more answers
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